$f(r, \theta) = \theta e^{-r^2}$ What is $\dfrac{\partial f}{\partial r}$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $e^{-r^2}$ (Choice B) B $e^{-r^2} - 2r\theta e^{-r^2}$ (Choice C) C $-2r e^{-r^2}$ (Choice D) D $-2r \theta e^{-r^2}$
Taking a partial derivative with respect to $r$ means treating $\theta$ like a constant, then taking a normal derivative. We can use the chain rule to differentiate $e^{-r^2}$, noting that $\theta$ acts like a constant here. $\begin{aligned} \dfrac{\partial f}{\partial r} &= \dfrac{\partial}{\partial r} \left[ \theta e^{-{r^2}} \right] \\ \\ &= \theta e^{-{r^2}} (-{2r}) \\ \\ &= -{2r} \theta e^{-{r^2}} \end{aligned}$ In conclusion, $\dfrac{\partial f}{\partial r} = -2r \theta e^{-r^2}$